550 AP Calculus AB & BC Practice Questions

Don’t simplify! Just plug in for x.

Take the derivative using the quotient rule.
Don’t simplify! Plug in 2 for x.

Find the derivative using the quotient rule in combination with the chain
rule.

Find the first derivative using the product rule.
f′(x) = (3x2)(cos x) + (x3)(–sin x)
= 3x2 cos x – x3 sin x
Find the derivative using the quotient rule and trigonometric derivatives.

y′ =

Find the second derivative.
f′(θ) = sin θ + θ cos θ
f″(θ) = cos θ + cos θ – θ sin θ
= 2 cos θ – θ sin θ
If y = x2 tan x, then y′ =
Take the derivative using trigonometric derivatives and the product rule.
y′ = (2x) tan x + (x2) sec2x.

Take the derivative using the rule of raising a number to a function.
f′(x) = 6x + 3 ln 6.
Take the derivative using the quotient rule.

Take the derivative using the derivative of the natural log function and
the chain rule.

Take the derivative using the product rule, chain rule, and trigonometric
derivatives.
y′ = 3x2 sin2x + x3 (2 sin x cos x)
Use the trigonometric identity sin 2x = 2 sin x cos x to rewrite the
derivative.
y′ = 3x2 sin2x + x3 sin 2x.
Although the product rule could be used to evaluate the derivative, this

question is quickly solved if you FOIL out the function first.

(x – 2)(2x + 3) = 2x2 – x – 6

Find the derivative.

f(x) = 2x2 – x – 6
f′(x) = 4x – 1

Calculate the derivative using the power rule for each term.

y = x6 – 3x4 + x

y′ = 3x5 – 12x3 + 1

Calculate the first derivative using the trigonometric derivatives and the
chain rule.

f(x) = sin 3x + 2 cos x – sin 2x
f′(x) = 3 cos 3x – 2 sin x – 2 cos 2x

Now take the second derivative paying attention to the sign changes
when applying the trigonometric derivatives.

f′(x) = 3 cos 3x – 2 sin x – 2 cos 2x
f″(x) = –9 sin 3x – 2 cos x + 4 sin 2x
f″(x) = 4 sin 2x – 9 sin 3x – 2 cos x

Take the derivative with respect to x.

Take the derivative with respect to x using the power rule and product
rule.

Take the derivative with respect to x using the power rule and product
rule.

Take the derivative with respect to x using the product rule, the chain
rule, and the differentiation formula for the sine function.

(2(x2 + 3)3) = 2((x2 + 3)2(2x)) = 12x(x2 + 3)2

Chapter 9

Derivatives Drill 2

DERIVATIVES DRILL 2

If f(x) = sin (csc x), then f′(x) =
(A) csc x cot x cos (csc x)
(B) –csc x cot x cos (csc x)
(C) cot2 x cos (csc x)
(D) –cot2 x cos (csc x)
(E) cos (–csc x cot x)

If y = 6x5 – 5x4, then y(4) =
(A) 30x4 – 20x3
(B) 120x3 – 60x2
(C) 360x2 – 120x
(D) 720x – 120
(E) 720

Find the derivative of f(x) = e2x sin2 3x.
(A) 2e2x sin 3x (sin 3x + 3cos 3x)
(B) 6e2x sin2 3x + 2e2x sin 3x cos x
(C) 2e2x sin 3x (sin 3x – 3cos 3x)
(D) –2e2x sin 3x (sin 3x + 3cos 3x)
(E) 2e2x sin 3x cos x – 6e2x sin2 3x

If y = ex (ln x)2, then y′ =
(A) 2ex ln x

ex ln x

ex ln x

(D) 2xex ln x

If y = tan2 (3θ), then y′ =
(A) sec2 (3θ) – 1
(B) 2 tan (3θ)
(C) 6 sec2 (3θ)
(D) 6 tan (3θ) sec2 (3θ)
(E) 3 tan (3θ) sec2 (3θ)

If f(x) = sin (sin (sin x)), then what is f′(x)?
(A) cos (cos (cos x))
(B) cos x cos (sin x) cos (sin (sin x))
(C) cos (sin (cos x))
(D) cos x + sin x cos x + sin2 x
(E) cos (cos x) cos x

If y = cos4 (sin3 x), then y′ =
(A) –12 sin2 x cos x sin (sin3 x) cos3 (sin3 x)
(B) 4 cos3 (sin3 x)
(C) 4 cos3 (3 sin2 x)
(D) cos4 (3 sin2 x)
(E) 12 sin2 x cos x sin (sin3 x) cos3 (sin3 x)

What is the derivative of f(x) = sin2 πx?
(A) 2π sin 2πx
(B) 2 cos πx

(C) sin 2πx
π sin 2πx

(E) sin πx cos πx

If y = cos (tan x), then y′ =
(A) sec2 x sin (tan x)
(B) –sin (sec2 x)
(C) –sec2 x sin (tan x)
(D) sec x tan x sin (tan x)
(E) –sec x tan x sin (tan x)

If f(x) = (x–2 + x–3)(x5 – 2x2), then f′(x) =
(A) –3x2 + 2x + 2x–2
(B) 3x2 + 2x – 2x–2

x3 + x2 – 2 – 2x–1
x4 + x3 – 2x + 2 ln x
(E) –3x2 – 2x – 2x2

What is the derivative of y = ?

Find the derivative of y =

If y = sin x, then y′ =
[sin x + 2x cos x]
[sin x + 2x cos x]
cos x

cos x
[sin x + 2x cos x]

If f(x) = x2 sin x tan x, then f′(x) =
(A) 2x cos x sec2x

x2 (cos x tan x + sin x sec2 x) + 2x (sin x tan x)
(C) –x2 (cos x tan x + sin x sec2 x) + 2x (sin x tan x)
(D) –2x cos x sec2 x

x2 (cos x + sec2 x)

What is the derivative of y = tan (sin x)?
(A) sec2 (cos x)
(B) cos x sec (sin x) tan (sin x)
(C) cos x sec2 (sin x)
(D) – cos x sec2 (sin x)
(E) – sec2 (cos x)

If y = 2 csc (4x), then y′ =
(A) –2 cot2 (4x)

(B) 2 cot2 (4x)
(C) –8 csc (x) cot (x)
(D) 2θ csc (4x) cot (4x)
(E) –8 csc (4x) cot (4x)

What is the derivative of y = (x2 + 2x)ex?
(A) (x + 1)ex
(B) 2(x + 1)ex
(C) 2x(x + 1)ex–1
(D) (x2 + 4x + 2)ex

x2 – 4x – 2)ex

If y = e2 ln x, then y′ =
(A) 0

(D) 2e ln x +

(E) 2e ln x −

Find f′(x) if f(x) = tan x + sec2 x.
(A) sec2 x (1 + 2 tan x)
(B) 2 sec2 x + tan x
(C) sec2 x (1 – 2 tan x)
(D) tan x – sec2 x
(E) 2 sec2 x – tan x

If y = esin x, then y′(π) =

(A) –2
(B) –1
(C) 0
(D) 1
(E) 2

Find y′ if y = ln (4x2 – 3x + 3)

What is the derivative of f(x) = x4 tan 5x?
x5 sec2 5x

(B) 20 x3 sec2 5x
x3 (4 tan 5x – 5x sec2 5x)

(D) –x3 (4 tan 5x – 5x sec2 5x)
x3 (4 tan 5x + 5x sec2 5x)
If y = tan2 (sin θ), then y′ =

(A) –2 cos θ tan (sin θ) sec2 (sin θ)
(B) 2 sec2 (cos θ)
(C) 2 tan (– cos θ)
(D) cos θ tan (sin θ) sec2 (sin θ)
(E) 2 cos θ tan (sin θ) sec2 (sin θ)

If y = sec (1 + x2), then y′ =

(A) sec (2x) tan (2x)
(B) 2x sec (2x) tan (2x)
(C) 2x sec2 (1 + x2)
(D) 2x sec (1 + x2) tan (1 + x2)
(E) 2x tan2 (1+ x2) sec (1 + x2)

If y sin (x2) = x sin (y2), then y′ =
(A) 2 sec (y2) – 2 csc (x2)

(C) 2 cos y – 2 cos x

If x2 cos y + sin 2y = xy, then find y′.

x + x2 sin y – 2 cos 2y
(C) 2x cos y – y

If sec y = sec2 x, then y′ =

(C) 2 sec2 x tan x
(D) tan y – 2 tan x
(E) sec y tan y

Chapter 10

Derivatives Drill 2 Answers and Explanations

ANSWER KEY

1. B
2. D
3. A
4. C
5. D
6. B
7. A
8. D
9. C
10. A
11. B
12. E
13. A
14. B
15. C
16. E
17. D
18. B
19. A
20. B
21. C
22. E
23. E
24. D
25. B
26. A
27. A

EXPLANATIONS

Take the derivative using trigonometric derivatives and the chain rule.

f(x) = sin (csc x)
f′(x) = cos (csc x)(– csc x cot x)
f′(x) = –csc x cot x cos (csc x)

Start calculating the derivatives to get to the 4th derivative by using the
power rule.

y = 6x5 – 5x4
y′ = 30x4 – 20x3
y″ = 120x3 – 60x2
y′(3) = 360x2 – 120x
y′(4) = 720x – 120

Take the derivative using the derivative of the exponential function,
trigonometric derivatives, the product rule, and the chain rule.

f(x) = e2x sin2 3x
f′(x) = (2e2x)(sin2 3x) + (e2x)(2 sin 3x cos 3x × 3)
f′(x) = 2e2x sin2 3x + 6e2x sin 3x cos 3x
f′(x) = 2e2x sin 3x (sin 3x + 3 cos 3x)

Take the derivative using the differentiation rules of exponential
functions, natural logarithms, the product rule and the chain rule.

y = ex (ln x)2

y′ = ex (ln x)2 + ex

y′ = ex (ln x)

Take the derivative using the trigonometric derivative formula for
tangent and the chain rule.

y = tan2 (3θ)
y′ = 2 tan (3θ) sec2 (3θ) 3
y′ = 6 tan (3θ) sec2 (3θ)

Take the derivative using the trigonometric formula for the sine function,
as well as the chain rule, starting from the outside and working your
way in the parentheses.

f(x) = sin (sin (sin x))
f′(x) = cos (sin (sin x)) cos (sin x) cos x
f′(x) = cos x cos (sin x) cos (sin (sin x))

Take the derivative using the trigonometric formula for the sine and
cosine functions and the chain rule.

y = cos4 (sin3 x)
y′ = 4 cos3 (sin3 x)(– sin (sin3 x)) 3 sin2 x cos x
y′ = –12 sin2 x cos x sin (sin3 x) cos3 (sin3 x)

Take the derivative using the trigonometric formula for the sine function
and the chain rule.

f(x) = sin2 πx
f′(x) = 2 sin πx cos πx π
f′(x) = 2π sin πx cos πx
f′(x) = π sin 2πx (* Note: sin 2x = 2 sin x cos x)

Take the derivative using the trigonometric formula for the cosine and
tangent functions, as well as the chain rule.

y = cos (tan x)
y′ = –sin (tan x) sec2x
y′ = –sec2x sin (tan x)
Although the product rule can be used to get this derivative, first apply
FOIL and then use the power rule of exponents to make it easier.
(x–2 + x–3)(x5 – 2x2) = x3 – 2 + x2 – 2x–1
f(x) = x3 + x2 – 2 – 2x–1
f′(x) = 3x2 + 2x + 2x–2
Take the derivative using the quotient rule.

Take the derivative using the quotient rule.

Take the derivative using the trigonometric formula for the sine function
and the product rule.

Take the derivative by carefully using the product rule, as well as the
trigonometric formula for both the sine and tangent functions.
f(x) = x2 sin x tan x
f′(x) = (2x)(sin x tan x) + (x2)(cos x tan x + sin x sec2 x)
Take the derivative of the function using the differentiation rules for the
tangent and sine functions, as well as the chain rule.
y = tan (sin x)

y′ = sec2(sin x) cos x
y′ = cos x sec2(sin x)

Take the derivative of the function using the differentiation rule for the
cosecant function, as well as the chain rule.

y = 2 csc (4x)
y′ = –8 csc (4x) cot (4x)

Take the derivative of the function using the product rule, the power
rule, and the exponential function rule.

y = (x2 + 2x)ex
y′ = (2x + 2)(ex) + (x2 + 2x)(ex)
y′ = (x2 + 4x + 2)ex

Take the derivative of the function using the constant rule, and the
derivative of the natural logarithm function. (Remember, e2 is a
constant!)

y = e2 ln x

y′ = e2

y′ =

Take the derivative using the rules of differentiation for both the tangent
and secant functions, as well as the power rule.

f(x) = tan x + sec2 x
f′(x) = sec2 x + 2 sec x sec x tan x
f′(x) = sec2 x (1 + 2 tan x)

Take the first derivative using the rule of exponential functions and of

the sine function.
y = e sin x
y′ = e sin x cos x
Do not simplify! Plug in x = π immediately.
y′ = e sin π cos π
= e0 (–1)
= –1
Take the derivative of the function by using the differentiation formula
for the natural logarithm function and the chain rule.
y = ln (4x2 − 3x + 3)

Take the derivative using the product rule, the differentiation formula for
the tangent function and the chain rule.
f(x) = x4 tan 5x
f′(x) = (4x3)(tan 5x) + (x4)(sec2 5x)(5)
f′(x) = x3 (4 tan 5x + 5x sec2 5x)
To find this derivative, you must utilize the chain rule and the rules of
trigonometric differentiation for the tangent and sine functions.
y = tan2 (sin θ)
y′ = 2 tan (sin θ) × sec2 (sin θ) × cos θ
So, y′ = 2 cos θ tan (sin θ) sec2 (sin θ)

Take the derivative using the trigonometric rule for the secant function
and the chain rule.
y = sec (1 + x2)
y′ = sec (1 + x2) tan (1 + x2) × 2x
y′ = 2x sec (1 + x2) tan (1 + x2)
Take the derivative with respect to x using the product rule, the chain
rule, and the differentiation formula for the sine function.

Take the derivative with respect to x using the product rule, chain rule,
and the differentiation rules for the sine and cosine functions.

Take the derivative with respect to x using the chain rule and the
differentiation rule for the secant function.



Chapter 11

Derivatives Drill 3

DERIVATIVES DRILL 3

Find y′ if x6 + y6 = 6.

(D) −

(E) −

Given h(x) = f (g(sin x)), what is h′ in terms of f′ and g′?
(A) cos x × f′(g(sin x)) × g′(sin x)

′(g′(cos x))
(g′(sin x)) × f′(g(sin x))
(D) cos x × f(g′(sin x)) × f′(g(sin x))
(E) –cos x × f′(g(sin x)) × g′(sin x)
If f(x) = x2 g(x), g(3) = 2, and g′(3) = 1, then what is f′(3)?
(A) 30
(B) 21
(C) 3
(D) –21
(E) –30
If y = 4x – tan x, then y′ =
(A) 4 + sec2 x

(B) 4 – sec x tan x
(C) 4 – sec2 x
(D) 4 + sec x tan x
(E) 4 sec2 x

If y = cot (3x2 + 5), then y′ =
(A) –csc2 (6x)
(B) –6x csc2 (3x2 + 5)
(C) 6x csc2 (3x2 + 5)
(D) –6x csc (3x2 + 5) cot (3x2 + 5)
(E) 6x csc (3x2 + 5) cot (3x2 + 5)

If y = csc (cot x), then y′ =
(A) –csc (csc2 x) cot (csc2 x)
(B) –csc2 x csc (cot x) cot (cot x)
(C) –csc2 x cot2 (cot x)
(D) csc2 x cot2 (cot x)
(E) csc2 x csc (cot x) cot (cot x)

If y = tan (sec x), then y′ =
(A) sec2 (sec x tan x)
(B) sec x tan x sec2 (sec x)
(C) sec (sec x) tan (sec x)
(D) –sec x tan x sec2 (sec x)
(E) –sec (sec x) tan (sec x)

Let r(x) = f(g(h(x))), where h(1) = 2, g(2) = 3, h′(1) = 4, g′(2) = 5,
and f′(3) = 6. What is the value of r′(1)?

(A) 1
(B) 3
(C) 20
(D) 120
(E) 720

If y = cot (csc x), then y′ =
(A) csc2 (csc x cot x)
(B) –csc2 (–csc x cot x)
(C) csc x cot x csc2 (csc x)
(D) –csc x cot x csc2 (csc x)
(E) csc2 (csc x) cot (csc x)

If y = sec (tan x), then y′ =
(A) sec (sec2 x) tan (sec2 x)
(B) tan2 (sec2 x) sec (tan x)
(C) –sec2 x sec (tan x) tan (tan x)
(D) sec2 x sec (tan x) tan (tan x)
(E) –sec (sec2 x) tan (sec2 x)

If F(x) = f (3 f (4 f (x))), where f(0) = 0 and f′(0) = 2, then what is the
value of F′(0)?
(A) 96
(B) 72
(C) 24
(D) 2
(E) 0

If x3 + y3 = 1, then y′ =

(C) –

If x2 + xy – y2 = 4, then y′ =

Find y′ if x3 + y3 = 6xy.

Find y′ if 4cos x sin y = 1.
(A) cot x cot y
(B) tan x cot y
(C) cot x tan y
(D) 1 – cos x sin y
(E) tan x tan y

If x3 + y3 = xy, then what is y′?
Find y′ if x = ln (x2 + y2).
x4 + y4 = π4. Find .
If 3x2 sin y = tan x, then y′ =
(E) 6x cos y – sec2 x

Find the derivative of y, when y = ?

of y = =

Find if y3 + 2y2 = 4x − 12.

Find if y3 + 2y2 = 4x − 12 and y = 1 at x = 7.

What is if y = log3(4x3 − 2x)?

Find the derivative of the inverse of y = x4 − 3 when y = –2.

(D) 1
Find for 4x2 − 2x2y + 2xy2 − 3y2 = x at x = 1.

(A) –4
(B) 0
(D) 1

If f(x) = 2x2 − 3x + 6, find a derivative of f−1(x) at y = 15.

(A) .
(B) −
(C)
(D) 3
(E) −

Find if y =

Chapter 12

Derivatives Drill 3 Answers and Explanations

ANSWER KEY

1. D
2. A
3. B
4. C
5. B
6. E
7. B
8. D
9. C
10. D
11. A
12. C
13. D
14. B
15. E
16. C
17. C
18. A
19. B
20. E
21. A
22. B
23. A
24. B
25. B
26. E
27. A
28. D

EXPLANATIONS

Take the derivative implicitly with respect to x and isolate y′.

You must find the derivative by utilizing the chain rule twice, as well as
the trigonometric differentiation formula for the sine function.
h(x) = f(g(sin x))
h′(x) = f′(g(sin x)) × g′(sin x) × cos x
h′(x) = cos x × f′(g(sin x)) × g′(sin x)
You must take the derivative using the power and product rules first.
f(x) = x2 g(x)
f′(x) = 2x g(x) + x2g′(x)
To find what f′(3) is equal to, you must use the other pieces of
information and plug in accordingly.
f′(3) = 2(3) g(3) + (3)2g′(3) = (6)(2) + (9)(1) = 21.
Take the derivative using the power rule and the trigonometric rule for

the tangent function.
y = 4x – tan x
y′ = 4 – sec2x

Take the derivative using the trigonometric rule for the cotangent
function and the chain rule.
y = cot (3x2 + 5)
y′ = –csc2 (3x2 + 5) × 6x
y′ = –6x csc2 (3x2 + 5)

Take the derivative using the trigonometric rules for the cosecant and
cotangent functions, as well as the chain rule.
y = csc (cot x)
y′ = –csc (cot x) cot (cot x) × –csc2x
y′ = csc2x csc (cot x) cot (cot x)

Take the derivative using the trigonometric differentiation rules for the
tangent and secant functions, as well as the chain rule.
y = tan (sec x)
y′ = sec2 (sec x) × sec x tan x
y′ = sec x tan x sec2 (sec x)

First, you must take the derivative using the chain rule twice.
r(x) = f(g(h(x)))
r′(x) = f′(g(h(x))) × g′(h(x)) × h′(x)

Evaluate the derivative at x = 1 using the information provided in the
question.
r′(x) = f′(g(h(x))) × g′(h(x)) × h′(x)
r′(1) = f′(g(h(1))) × g′(h(1)) × h′(1)
r′(1) = f′(g(2)) × g′(2) × (4)
r′(1) = f′(3) × (5) × (4)
r′(1) = (6) × (5) × (4) = 120

Take the derivative using the trigonometric differentiation rules for the
cotangent and cosecant functions, as well as the chain rule.
y = cot (csc x)
y′ = –csc2 (csc x)(–csc x cot x)
y′ = csc x cot x csc2 (csc x)

Take the derivative using the trigonometric differentiation rules for the
secant and tangent functions, as well as the chain rule.
y = sec (tan x)
y′= sec (tan x) tan (tan x) × sec2 x
y′ = sec2 x sec (tan x) tan (tan x)

Take the derivative of F(x) using the chain rule.
F(x) = f (3 f (4 f (x)))
F′(x) = f′(3 f (4 f (x))) × 3 f′(4f(x)) × 4 f′(x)
Evaluate the derivative at x = 0 using the information provided in the
question.

F′(x) = f′(3 f (4 f (x))) × 3 f′(4f (x)) × 4 f′(x)
F′(0) = f′(3 f (4 f (0))) × 3 f′(4f (0)) × 4 f′(0)
F′(0) = f′(3 f (4 (0))) × 3 f′(4(0)) × 4(2)
F′(0) = f′(0) × 3(2) × 4(2)
F′(0) = (2) × 3(2) × 4(2) = 96
Calculate the derivative with respect to x.

Calculate the derivative with respect to x.

Calculate the derivative with respect to x.

Calculate the derivative with respect to x.

Calculate the derivative with respect to x.
Calculate the derivative with respect to x.

Calculate the derivative with respect to x.
Calculate the derivative with respect to x.

This problem can be solved using the Chain rule, Product rule, and
Quotient rule, but that can get especially messy. We are going to use
logarithmic differentiation to solve it. First, take the natural log of both

sides: ln y = ln . Use logarithmic rules to simplify

the equation: ln y = ln(x3 − 2x2) + 2 ln(sin x) − 3 ln(x2 + 1). Now,
differentiate both sides with respect to x:

. Finally, isolate :

.

Use the chain and quotient rules:

.

Use implicit differentiation: . Thus,
Via implicit differentiation, and
. Evaluate at (7, 1), to get,

. Plug in these points to solve

Recall, . In this problem, u = 4x2 − 2x and du =
(12x2 − 2)dx. Thus,

.

. Here, y = –2 when x = 1 and

. Evaluate this derivative at x = 1 to get the solution:

.

First, plug x = 1 into the equation and solve for y. The result will be two
y values, y = –3 and y = 1. Next, use implicit differentiation to find the

first derivative: 8x − 2x2 − 4xy + 4xy + 2y2 − 6y = 1.

Do not simplify because the questions asks for the derivative at x = 1.
Thus, the final step is to plug (1,1) and (1, –3) into the equation for the
first derivative and solve for the derivatives at these points. At (1,1),

. At (1, –3), . Since the derivative at (1, –3) is an

option, the answer is E.

The derivative of the inverse is: . To

begin, determine the x-value that corresponds to y = 15. There are two
possible x-values: x = − and x = 3. Next, find the derivative of the

function f(x): f(x) = 4x − 3. Plug this into the left hand side of the
formula by first taking the inverse. Then, evaluate the function at both x-

values: and .

Because the quotient rule and chain rule would get messy here, use
logarithmic differentiation. First, take the natural log of both sides of the

equation and simplify: ln y = ln = 2 ln(3x3 + 2) − ln(x

− 2). Now, differentiate both sides using implicit differentiation:

. Solve for : .

Chapter 13

Applications of Derivatives Drill 1

APPLICATIONS OF DERIVATIVES DRILL
1

Find the equation of the line tangent to y = at (1,1).

x – 2y = –1
(B) 2x – y = 1

x – y = 2
x – 2y = 1
(E) 2x + y = 1

If f(x) = , then find f″(2).
(A) –8
(B) –

(C) 0

(E) 8

Find the critical numbers of y = 3x4 + 4x3 – 12x2.
(A) 0
(B) –2, 1
(C) 0, 1
(D) –2, 0, 1
(E) –1, 0, 2

What is the maximum value of f(x) = 2x3 – 3x2 – 12x + 1 on the
interval [–2,3]?
(A) –3
(B) 0
(C) 2
(D) 6
(E) 8

Find the interval(s) on which f is decreasing for f(x) = 2x3 + 3x2 – 36x.
(A) (–∞,–3)
(B) (–2,3)
(C) (–3,2)
(D) (2,3)
(E) (2,∞)

Find all critical numbers of y = 2x3 – 3x2 – 12x.
(A) –2
(B) –1
(C) –2, –1
(D) –1, 2
(E) 1, 2

Find any points of inflection of y = x4 + 4x3.
(A) (–2,–16)
(B) (0,0)
(C) (2,16) and (0,0)
(D) (0,0) and (–2,–16)
(E) (2,16)

Find the equation of the line tangent to y = sin (sin x) at (π, 0).
(A) x – y = π

x + y = π
(C) 2x – y = π

x – 2y = π
x + y = 2π

Find the minimum value of f(x) = 2x3 + 3x2 – 36x.
(A) –44
(B) –9
(C) 3
(D) 9
(E) 44

What is the point of inflection of f(x) = (x + 1)5 – 5x – 2?
(A) (–3,1)
(B) (–1,3)
(C) (0,0)
(D) (1,3)
(E) (3,1)

On what interval(s) is f decreasing if f(x) = 2 + 2x2 – x4?
(A) (–1,0) only
(B) (1,∞) only
(C) (–∞,–1) and (0,1)
(D) (–1,0) and (1,∞)
(E) (0,1) only

A particle is traveling according to f(x) = x3 – 12x2 + 36x. What is the
velocity at x = 3 seconds?
(A) –18
(B) –9
(C) 0
(D) 9
(E) 18

If a ball is thrown upward with a velocity of 80 ft/s, then its height after
t seconds is s = 80t – 16t2. What is the maximum height of the ball?
(A) 2.5
(B) 80
(C) 100
(D) 180
(E) 270

Find the equation of the normal line to the curve y = at (1, ).

(A) 8x + 2y = 7
(B) 2x + 8y = 7
(C) 8x – 2y = 7
(D) 2x – 8y = 7
(E) 8x – 2y = –7

For what values of x does the graph of f(x) = x + 2sin x have a
horizontal tangent on [0,2π]?

and

and

and

only (E) no values

Find an equation of the tangent line to the curve y = 2x sin x at the
point ( , π).

y = 2x + π
y = 2x – π
y = 2x

y = 2x +

y = 2x −

A particle travels in a position governed by the equation s(t) = 4t3 –
16t2. What is its acceleration at t = 2 seconds?
(A) 0
(B) 2
(C) 10
(D) 12
(E) 16

If a particle travels along a path according to the equation s(t) = 6t2 – 4t
+ 3, then what is the velocity at t = 2 seconds?
(A) –20
(B) –10
(C) 0

(D) 10
(E) 20

Find the absolute maximum value of f(x) = (x2 + 2x)3 on the interval [–
2,1].
(A) –2
(B) –1
(C) 0
(D) 1
(E) 27

What is the x-coordinate of the point of inflection of f(x) = 4x3 + 3x2 –
6x?
(A) –4

(B) –

(C) 0

(E) 4 ?

On what interval(s) is f decreasing for f(x) =

(A) (–∞,∞)
(B) (–∞,0)
(C) (0,∞)
(D) (–∞,–3)
(E) (–3,∞)

Questions 22–23 rely on the following information: Suppose that h(x) =